Aptitude - Logarithm - Discussion
Discussion Forum : Logarithm - General Questions (Q.No. 5)
5.
| If log | a | + | log | b | = log (a + b), then: |
| b | a |
Answer: Option
Explanation:
| log | a | + log | b | = log (a + b) |
| b | a |
 log (a + b) = log |
 |
a | x | b |  |
= log 1. |
| b | a |
So, a + b = 1.
Discussion:
33 comments Page 3 of 4.
Shree said:
1 decade ago
What is the solution for:
log{(a+b)/3} = 0.5(log a + log b).
The answer is a^2+b^2 = 7ab.
But I want the procedure using the formula as mentioned here please.
log{(a+b)/3} = 0.5(log a + log b).
The answer is a^2+b^2 = 7ab.
But I want the procedure using the formula as mentioned here please.
Dilshan fernando said:
1 decade ago
log(a/b)+log(b/a) = log(a+b).
log(a/b*b/a) = log(a+b).
log 1 = log (a+b).
log 1 = 0.
So that a+b = 0.
a = -b.
log(a/b*b/a) = log(a+b).
log 1 = log (a+b).
log 1 = 0.
So that a+b = 0.
a = -b.
Rajaa said:
1 decade ago
Problem: log(a/b)+log(b/a) = log(a+b).
log(a)+log(b) = log(a*b).
So LHS log(a/b)+log(b/a) = log(a/b*b/a).
= log(1).
And RHS log(a+b).
LHS = RHS.
log(a+b) = log(1).
Hence a+b = 1.
log(a)+log(b) = log(a*b).
So LHS log(a/b)+log(b/a) = log(a/b*b/a).
= log(1).
And RHS log(a+b).
LHS = RHS.
log(a+b) = log(1).
Hence a+b = 1.
Manikantan said:
1 decade ago
log a/b +log b/a = log a - log b + log b - log a = 0.
log (a+b)=0 => a+b =1 *(log 1 =0).
log (a+b)=0 => a+b =1 *(log 1 =0).
Keshav said:
1 decade ago
@Sameer according to the logarithmic identity.
log A + log B = log(A*B).
& log A + log B = log(A+B).
There is no such identity.
Please check.
log A + log B = log(A*B).
& log A + log B = log(A+B).
There is no such identity.
Please check.
Sameer said:
1 decade ago
Here someone wrote: loga+logb = log(a+b).
But how log a/b + log b/a = log (a/b x b/a)?
Why not like this log(a/b + b/a)?
But how log a/b + log b/a = log (a/b x b/a)?
Why not like this log(a/b + b/a)?
MADHU said:
1 decade ago
Can any one explain how log(a+b)=loga+logb
Raghavendra said:
1 decade ago
log(a+b)=log(a/b)+log(b/a)
log(a+b)=l0ga-logb+logb-loga
log(a+b)=0
a+b=10^0
so
a+b=1
log(a+b)=l0ga-logb+logb-loga
log(a+b)=0
a+b=10^0
so
a+b=1
Priya said:
1 decade ago
Thanks siddhu.
Siddhu said:
1 decade ago
We know that: log a + log b = log (a+b);
Similarly log(a/b) + log(b/a) = log (a/b * b/a);
We get log 1 bcoz all terms will get cancels
and log 1 = 0;
This is about L.H.S. So look at the R.H.S log(a+b) is there and we have to get R.H.S also zero it is possible when we get log 1 = log(a+b) so a+b = 1.
Similarly log(a/b) + log(b/a) = log (a/b * b/a);
We get log 1 bcoz all terms will get cancels
and log 1 = 0;
This is about L.H.S. So look at the R.H.S log(a+b) is there and we have to get R.H.S also zero it is possible when we get log 1 = log(a+b) so a+b = 1.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers